A line passes through (-1, -3) and is perpend | vedclass

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(B) The given line is $x + 6y = 5$,which can be written as $y = -\frac{1}{6}x + \frac{5}{6}$.The slope of this line is $m_1 = -\frac{1}{6}$.Since the

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$-\frac{1}{2}$

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(B) The given line is $x + 6y = 5$,which can be written as $y = -\frac{1}{6}x + \frac{5}{6}$.The slope of this line is $m_1 = -\frac{1}{6}$.Since the required line is perpendicular to this line,its slope $m_2$ must satisfy $m_1 	imes m_2 = -1$.Thus,$m_2 = -\frac{1}{m_1} = -\frac{1}{-1/6} = 6$.The equation of the line passing through $(-1, -3)$ with slope $m = 6$ is given by $y - y_1 = m(x - x_1)$.$y - (-3) = 6(x - (-1))$$y + 3 = 6(x + 1)$$y + 3 = 6x + 6$$6x - y = -3$.To find the $x$-intercept,we set $y = 0$ in the equation:$6x - 0 = -3$$6x = -3$$x = -\frac{3}{6} = -\frac{1}{2}$.

$A$ line passes through $(-1, -3)$ and is perpendicular to $x + 6y = 5$. Its $x$-intercept is

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